Example 1 (Asymptotes)

Let

f(x)=\frac{3x^2-2x+2}{x^2-x-2}

What is the domain of f? What are the roots and y-intercepts of f? Find the horizontal and vertical asymptotes of f.

Solution The domain of f is the set on which f is defined. As a quotient of continuous functions, f is defined when the denominator is not zero:

x^2-x-2\neq0,

\Rightarrow (x-2)(x+1)\neq 0.

That is the domain of f is \mathbb{R}\backslash\{-1,2\} (all the real numbers except -1 and 2. Note now that

f(x)=\frac{3x^2-2x+2}{(x-2)(x+1)}).

The roots of a quotient of continuous functions occur at points in the domain when the numerator is zero:

3x^2-2x+2=0,

\Rightarrow x_{\pm}=\frac{2\pm \sqrt{4-4(3)(2)}}{6}=\frac{2\pm\sqrt{-20}}{6}\not\in\mathbb{R}.

That is f has no (real) roots (\sqrt{-20} is not a real number).

 

The y-intercept of f is the point where f cuts the y-axis (i.e. when x=0):

f(0)=\frac{3(0)^2-2(0)+2}{(0)^2-(0)-2}=\frac{2}{-2}=-1.

 

The horizontal asymptotes of f are the behaviours of f as x\rightarrow \pm\infty:

\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow \infty}\frac{3x^2-2x+2}{x^2-x-2}\approx \frac{3x^2}{x^2}=3.

Hence h(x)=3 is the horizontal asymptote.

 

The vertical asymptotes of f are the finite x where f gets very big (has a singularity). Following our procedure:

Step 1: A quotient of continuous functions can only have a vertical asymptote when the denominator is zero; i.e. x=-1,2 (it is necessary but not sufficient that the denominator is zero – hence Step 2).

Step 2: Evaluate the limits

\lim_{x\rightarrow-1}f(x) and \lim_{x\rightarrow 2}f(x)

\lim_{x\rightarrow -1}\frac{3x^2-2x+2}{x-2}.\frac{1}{x+1}=\frac{3(-1)^2-2(-1)+2}{-1-2}.\lim_{x\rightarrow -1}\frac{1}{x+1}=\frac{7}{-3}\times\pm\infty=\pm\infty.

\lim_{x\rightarrow 2}\frac{3x^2-2x+2}{x+1}.\frac{1}{x-2}=\frac{3(2)^2-2(2)+2}{2+1}.\lim_{x\rightarrow 2}\frac{1}{x-2}=\frac{10}{3}\times\pm\infty=\pm\infty.

Step 3: As these limits evaluate to infinity there are vertical asymptotes at x=-1 and x=2.

Comment To carry out the full analysis of this function to sketch it is quite messy – we need a computer to find the split points of f' and f''.

 

Example 2 (all elements considered)

Let

f(x)=\frac{x^2+6x-4}{3x-2}

Find the domain of f. Find the roots and y-intersections of f. Find the horizontal and vertical asymptotes of f. Find the intervals where f is increasing/ decreasing. Find the locations of the local extrema (local max/min). Find where the function is concave up/ concave down. Hence sketch the graph.

Solution The domain of f is where

3x-2\neq 0\Leftrightarrow 3x\neq 2

\Leftrightarrow x\neq\frac{2}{3}

Domain of f is \mathbb{R}\backslash\{2/3\}.

The roots of f are where

x^2+6x-4=0

x_{\pm}=\frac{-6\pm \sqrt{36-4(-4)}}{2}=\frac{-6\pm\sqrt{4(13)}}{2}=-3\pm\sqrt{13}.

The y -intercept, f(0)=\frac{-4}{-2}=2.

The horizontal asymptotes

\lim_{x\rightarrow\infty}\frac{x^2+6x-4}{3x-2}\approx \frac{x^2}{3x}=\frac{1}{3}x.

Hence the horizontal asymptote h(x)=x/3(Which is a line of slope 1/3 and y-intercept 0. (Cf. y=mx+c)).

The vertical asymptotes:

Step 1: x=2/3.

Step 2:

\lim_{x\rightarrow 2/3}f(x)= ((2/3)^2+6(2/3)-4).\lim_{x\rightarrow 2/3}\frac{1}{3x-2}=\frac{4}{9}\times\pm\infty=\pm\infty

Step 3: Hence x=2/3 is a vertical asymptote.

To find where f is increasing/ decreasing, we have to find the split points of f' (the only place f' can change sign; i.e. the only place where f can go from increasing to decreasing or vice versa . Remember f'>0\Rightarrow increasing; f'<0\Rightarrow decreasing. The split points are where f'=0 or undefined). By the quotient rule, f' is undefined when the denominator is zero; i.e. x=2/3. By the quotient rule:

f'(x)=\frac{(3x-2)(2x+6)-(x^2+6x-4)(3)}{(3x-2)^2}

=\frac{6x^2+18x-4x-12-3x^2-18x+12}{(3x-2)^2}=\frac{3x^2-4x}{(3x-2)^2}=\frac{x(3x-4)}{(3x-2)^2}

The other split points are where f'=0, and in this case this occurs when x(3x-4)=0; i.e. x=0,4/3. Now we test in between the split points to see if f' is positive/negative (f increasing/decreasing). The split points are 0,2/3,4/3 so the intervals to test are:

(-\infty,0),(0,2/3),(2/3,4/3),(4/3,\infty)

Choose test points -1,1/2,1,2:

f'(-1)=\frac{(-1)(3(-1)-4)}{(3(-1)-2)^2}=\frac{(-)(-)}{(+)}>0

f'(1/2)=\frac{(+)(-)}{(+)}<0

f'(1)=\frac{(+)(-)}{(+)}<0

f'(2)=\frac{(+)(+)}{(+)}>0

Hence f'>0 on (-\infty,0)\cup(4/3,\infty) and f'<0 on (0,2/3)\cup(2/3,4/3); i.e. increasing on (-\infty,0)\cup(4/3,\infty) and decreasing on (0,2/3)\cup(2/3,4/3).

To find out the locations of the local extrema, we do the following schematic of f (If f is increasing on an interval draw it increasing, if f is decreasing on an interval draw it decreasing. Beware of singularities! At these points f is not continuous (like it is at 0 and 4/3)).:

(please ignore the vertical line joining them up) It is clear from this schematic that there is a local maxima at 0 and a local minima at 4/3. These have coordinate (0,2) and (4/3,f(4/3))=(4/3,26/9).

To find where f is concave up/ concave down we find the split points of f''. By the quotient rule, as

f'(x)=\frac{x(3x-4)}{(3x-2)^2},

f'' is undefined at x=2/3. By the quotient rule:

f''(x)=\frac{(3x-2)^2(6x-4)-x(3x-4)2(3x-2)(3)}{(3x-2)^4}

Because we have already considered the case when x=2/3, we may cancel the 3x-2 (i.e. dividing by 3x-2 is not troublesome as we have already identified x=2/3 as a split point and hence we will not be dividing by zero):

f''(x)=\frac{(3x-2)(6x-4)-6x(3x-4)}{(3x-2)^3}

f''(x)=\frac{18x^2-12x-12x+8-18x^2+24x}{(3x-2)^3}=\frac{8}{(3x-2)^3}\neq 0,\,\forall\,x\in\mathbb{R}

Hence the only split point of f'' is x=2/3. Now we test in between the split points to see if f'' is positive/negative (f concave up/ down). The split point is 2/3 so the intervals to test are:

(-\infty,2/3),(2/3,\infty)

Choose test points 0,1:

f''(0)=\frac{8}{(3(0)-2)^3}=\frac{(+)}{(-)}<0

f''(1)=\frac{(+)}{(+)}>0

Hence f''>0 on (2/3,\infty) and f''<0 on (-\infty,2/3); i.e. concave up on on (2/3,\infty) and concave down on  (-\infty,2/3).

To sketch the graph take all these points into consideration (click for a sharper image):



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